C-language printing leap year 1000--2000

Leap Year Judgment method:

1. Years that can be divisible by 400;

2. Year that can be divisible by 4 but not divisible by 100 at the same time.

A leap year is one that satisfies either of these conditions.

Here are three different forms

The procedure is as follows:

One:

#include < stdio . h >

int Main ()

{

int =

for (M = <= ++ )

{if (M % == 0)

{if (M % == 0)

{if(M%= =0)

N = 1;

Else n=0;

        }

Else n=1;

     }

else =

if(n) printf ("%d", M,count+ +);

   }

if % == 0) printf ( );p rintf ( "\ncount=%d" ,count);

return 0;

}

Two:

#include < stdio . h >

int Main ()

{

int m,n,count=0;

for (M = <= ++ )

{if (M % == 0)

{if (M % != 0) printf ( "%d " ++ );

if (M % == 0)

printf ("%d", M,count+ +);

}printf ("\ncount=%d", count);

return 0;

}

Three:

#include < stdio . h >

int Main ()

{

int m,count=0;

for(m=1000;m<=2000;m+ +)

  {

  if(M%4==0&&m% -!=0||m% -==0) printf ("%d", M,count++);

  }

printf ("\ncount=%d", count);

return 0;

}

This is known as the "four Years a leap, a century does not leap, 400 years to leap" Leap year algorithm program. But for a large number of years, if the year can be divisible by 3200, and divisible by 172800 is a leap, this program is not considered.

This article is from the "Materfer" blog, make sure to keep this source http://10741357.blog.51cto.com/10731357/1698504

C-language printing leap year 1000--2000